Problem: $\overline{BC} = 10$ $\overline{AB} = {?}$ $A$ $C$ $B$ $?$ $10$ $ \sin( \angle BAC ) = \frac{10\sqrt{109} }{109}, \cos( \angle BAC ) = \frac{3\sqrt{109} }{109}, \tan( \angle BAC ) = \dfrac{10}{3}$
$\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AB}$ is the hypotenuse (note that it is opposite the right angle) SOH CAH TOA We know the opposite side and need to solve for the hypotenuse so we can use the sin function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}} = \frac{10}{\overline{AB}} $ $ \overline{AB}=\frac{10}{\sin( \angle BAC )} = \frac{10}{\frac{10\sqrt{109} }{109}} = \sqrt{109}$